Worked example 1
What volume does 2.50 mol of nitrogen occupy at 1.20 atm and 35 °C?
Try it first: Convert the temperature and rearrange the equation before you reach for a calculator.
0 of 4 steps revealed.
What you'll be able to do: Use PV = nRT with consistent units to find any one gas variable, and derive the simple gas laws from it.
One equation replaces every named gas law you may have memorised. Get the units right and PV = nRT will answer almost any single-state gas question.
These are recommended, not required. You can start this lesson at any time.
PV = nRT relates the four state variables of a gas sample. Use R = 0.08206 L*atm/(mol*K) when pressure is in atmospheres, or R = 8.314 J/(mol*K) when you need energy units. Temperature is always in kelvin, obtained by adding 273.15 to a Celsius value.
PV = nRT
Hold n and T constant and PV is a constant, which is the law of Boyle. Hold n and P constant and V/T is a constant, which is the law of Charles. Hold P and T constant and V/n is a constant, which is the law of Avogadro. You never need to memorise them separately.
When the same sample moves between two states, divide one form of the equation by the other. Anything held constant cancels, leaving P₁V₁/T₁ = P₂V₂/T₂. If moles also change, keep n in the expression as P₁V₁/(n₁T₁) = P₂V₂/(n₂T₂).
P₁V₁/T₁ = P₂V₂/T₂
Substituting n = m/M into PV = nRT gives PM = dRT, where d is density in grams per litre. This is how a gas density measurement identifies an unknown gas, and it explains why warm air, at lower density, rises.
At STP, defined as 273.15 K and 1 atm, one mole of ideal gas occupies 22.4 L.
Every gas law calculation uses absolute temperature; Celsius values give nonsense ratios.
PV = nRT describes one state of a sample, so a change needs the equation applied twice.
PM = dRT lets a density measurement give a molar mass.
PV = nRT
PM = dRT
What volume does 2.50 mol of nitrogen occupy at 1.20 atm and 35 °C?
Try it first: Convert the temperature and rearrange the equation before you reach for a calculator.
0 of 4 steps revealed.
A gas has a density of 1.96 g/L at 1.00 atm and 273 K. Identify its molar mass.
Try it first: Choose between PV = nRT and PM = dRT based on the data you were given.
0 of 4 steps revealed.
Why it's wrong: Ratios of Celsius values are not ratios of absolute temperature, so the answer can even be negative.
Check instead: Add 273.15 as the very first step of every gas problem.
Why it's wrong: The molar volume changes with both pressure and temperature.
Check instead: Use PV = nRT unless the problem states standard conditions.
Why it's wrong: The units of R must match the units substituted, or the result is off by a large factor.
Check instead: Convert pressure to atm, or switch to a value of R with matching units.
No practice questions are available for this topic yet. You can still practice the whole unit.
The ideal gas law links pressure, volume, moles and temperature through the gas constant R. With R = 0.08206 L*atm/(mol*K), pressure must be in atmospheres, volume in litres and temperature in kelvin. Holding two variables fixed recovers the laws of Boyle, Charles and Avogadro, and the two-state form P₁V₁/T₁ = P₂V₂/T₂ handles changes.
This lesson is original Chem Help content. No external sources were adapted.